Let `f(x)=(x+1)^2-1, xgeq-1.` Then the set `{x :f(x)=f^(-1)(x)}` is `{0,1,(-3+isqrt(3))/2,(-3-isqrt(3))/2}` (b) `{0,1,-1` `{0,1,1}` (d) `e m p t y`

Options

A. `{0,-1,(-3+isqrt(3))/(2),(-3-isqrt(3))/(2)}`
B. `{0,1,-1}`
C. `{0, -1}`
D. empty

Explanation

To solve the problem, we need to find the set of \( x \) such that \( f(x) = f^{-1}(x) \) for the function \( f(x) = (x+1)^2 - 1 \) where \( x \geq -1 \).

### Step 1: Find the function \( f(x) \)
The function is given by:
\[ f(x) = (x+1)^2 - 1 \]
Expanding this, we get:
\[ f(x) = x^2 + 2x + 1 - 1 = x^2 + 2x \]

### Step 2: Find the inverse function \( f^{-1}(x) \)
To find the inverse, we set \( y = f(x) \):
\[ y = x^2 + 2x \]
Rearranging gives:
\[ x^2 + 2x - y = 0 \]
This is a quadratic equation in \( x \). We can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 2, c = -y \):
\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-y)}}{2 \cdot 1} = \frac{-2 \pm \sqrt{4 + 4y}}{2} \]
Simplifying this, we have:
\[ x = \frac{-2 \pm 2\sqrt{1 + y}}{2} = -1 \pm \sqrt{1 + y} \]
Since \( f(x) \) is increasing for \( x \geq -1 \), we take the positive root:
\[ f^{-1}(y) = -1 + \sqrt{1 + y} \]

### Step 3: Set up the equation \( f(x) = f^{-1}(x) \)
We need to solve:
\[ f(x) = f^{-1}(x) \]
Substituting the expressions we found:
\[ x^2 + 2x = -1 + \sqrt{1 + x} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x^2 + 2x + 1 = \sqrt{1 + x} \]
Squaring both sides:
\[ (x^2 + 2x + 1)^2 = 1 + x \]
Expanding the left side:
\[ (x^2 + 2x + 1)(x^2 + 2x + 1) = x^4 + 4x^3 + 6x^2 + 4x + 1 \]
Setting this equal to \( 1 + x \):
\[ x^4 + 4x^3 + 6x^2 + 4x + 1 = 1 + x \]
Subtracting \( 1 + x \) from both sides:
\[ x^4 + 4x^3 + 6x^2 + 3x = 0 \]

### Step 5: Factor the equation
Factoring out \( x \):
\[ x(x^3 + 4x^2 + 6x + 3) = 0 \]
This gives us one solution:
\[ x = 0 \]
Now we need to solve the cubic equation \( x^3 + 4x^2 + 6x + 3 = 0 \).

### Step 6: Finding roots of the cubic equation
Using the Rational Root Theorem or synthetic division, we can find the roots. Testing \( x = -1 \):
\[ (-1)^3 + 4(-1)^2 + 6(-1) + 3 = -1 + 4 - 6 + 3 = 0 \]
So \( x = -1 \) is a root. We can factor:
\[ x^3 + 4x^2 + 6x + 3 = (x + 1)(x^2 + 3x + 3) \]
Now we solve \( x^2 + 3x + 3 = 0 \) using the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} = \frac{-3 \pm \sqrt{9 - 12}}{2} = \frac{-3 \pm \sqrt{-3}}{2} = \frac{-3 \pm i\sqrt{3}}{2} \]

### Step 7: Collecting all solutions
Thus, the solutions are:
1. \( x = 0 \)
2. \( x = -1 \)
3. \( x = \frac{-3 + i\sqrt{3}}{2} \)
4. \( x = \frac{-3 - i\sqrt{3}}{2} \)

### Final Result
The set of \( x \) such that \( f(x) = f^{-1}(x) \) is:
\[ \{0, -1, \frac{-3 + i\sqrt{3}}{2}, \frac{-3 - i\sqrt{3}}{2}\} \]